Question

$\underset{x\to 0}{lim}\frac{e^{x^3}-1-x^3}{{\sin }^6\left(2x\right)}$ equals

• A. $\frac{1}{128}$
• B. $\frac{2}{127}$
• C. $\frac{1}{126}$
• D. $\frac{1}{125}$

Answer 1

The correct option is A $\frac{1}{128}$

We know the expansion
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$
Putting $x\to x^3$
$\Rightarrow e^{x^3}=1+x^3+\frac{{\left(x^3\right)}^2}{2!}+\frac{{\left(x^3\right)}^3}{3!}+\cdots$

$\Rightarrow e^{x^3}-1-x^3=x^6[\frac{1}{2}+\frac{x^3}{6}+\cdots ]$

$\therefore \underset{x\to 0}{lim}\frac{e^{x^3}-1-x^3}{\frac{{\sin }^6\left(2x\right)}{(2x{)}^6}(2x{)}^6}$

$=\underset{x\to 0}{lim}\frac{x^6[\frac{1}{2}+\frac{x^3}{6}+\cdots ]}{\frac{{\sin }^6\left(2x\right)}{(2x{)}^6}\times (2x{)}^6}$

$=\underset{x\to 0}{lim}\frac{[\frac{1}{2}+\frac{x^3}{6}+\cdots ]}{\frac{{\sin }^6\left(2x\right)}{(2x{)}^6}\times 2^6}$
$=\frac{1}{2^7}=\frac{1}{128}(\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1)$